Weighted Average problem. Pretty easy, please help.?

Question

And show the work.

Here is the problem:
Charley wants to create a special mix of two coffees. One priced $6.40 per lb, the other at $7.28 per lb. How many pounds of the $7.28 coffee should he mix with 9 lbs of the 6.40 coffee to sell the mixture for $6.95 per pound?

Answer 1

(7.28x+57.6)/x+9=6.95
7.28x+57.6=6.95x+62.55
7.28x=6.95x+4.95
.33x=4.95
4.95/.33=x

15=x

Answer 2

Okay, just to make this easier:
x = 6.40
y = 7.28
z = number of pounds of coffee y in the recipe

(9x + zy) / (9+z) = 6.95 (That is, 9 pounds of x and z pounds of y will average to be $6.95 per pound)

9x + zy = 6.95*9 + 6.95*z
57.60 + z*7.28 = 62.55 + 6.95*z
57.60 + 0.33*z = 62.55
0.33z = 4.95
z = 15

Thus, you need 15 pounds of the $7.28 coffee to make the correct blend.

Answer 3

9 lbs * 640/lb+ x lbs * 728/lb = (9 + x) lbs * 695/lb
5760 + 728x = 6255 + 695x
33x = 495
x = 15 pounds

I always change dollars and cents into cents to work these problems. It avoid the decimal issues.

Answer 4

a = # pounds of $6.95/pound coffee
b = # pounds of $7.28/pound coffee

6.4(a) + 7.28(b) = total
6.4(9) + 7.28(b) = 6.95 (9+b)
57.6 + 7.28(b) = 62.55 + 6.95(b)
7.28(b) - 6.95(b) = 62.55 - 57.6
.33(b) = 4.95 DIVIDE both sides by .33
b = 4.95/.33
b= 15